元组scala的不可变迭代器映射
我在scala中有一个元组映射,我只想从中获得选中的键、值对 我试过了元组scala的不可变迭代器映射,scala,dictionary,iterator,tuples,immutability,Scala,Dictionary,Iterator,Tuples,Immutability,我在scala中有一个元组映射,我只想从中获得选中的键、值对 我试过了 val m1 = {"a":{"0":"R","1":null,"2":null,"3":12.25.0,"4":null} ,"b":{"0":"R","1":null,"2":null,"3":34.75,"4":null} ,"c":{"0":"R","1":null,"2":null,"3":56.25,"4":null} ,"d":{"0":"R","1":nu
val m1 = {"a":{"0":"R","1":null,"2":null,"3":12.25.0,"4":null}
,"b":{"0":"R","1":null,"2":null,"3":34.75,"4":null}
,"c":{"0":"R","1":null,"2":null,"3":56.25,"4":null}
,"d":{"0":"R","1":null,"2":null,"3":45.65,"4":null}
}
然后尝试迭代。但我无法处理这个复杂的表达式
因为我只需要a和b,结果会是
a-> 12.25
d-> 45.65
如何在scala中迭代它 数据的Scala等价物如下所示:
val m1: List[(String, List[(String, Any)])] =
List("a" -> List("0" -> "R", "1" -> null, "2" -> null, "3" -> 12.25, "4" -> null),
"b" -> List("0" -> "R", "1" -> null, "2" -> null, "3" -> 34.75, "4" -> null),
"c" -> List("0" -> "R", "1" -> null, "2" -> null, "3" -> 56.25, "4" -> null),
"d" -> List("0" -> "R", "1" -> null, "2" -> null, "3" -> 45.65, "4" -> null))
val keys = List("a", "d")
val field = 3
m1.collect{ case (k, v) if keys.contains(k) => (k, v(field)._2) }
获得此格式的数据后,可以按如下方式提取所需的数据:
val m1: List[(String, List[(String, Any)])] =
List("a" -> List("0" -> "R", "1" -> null, "2" -> null, "3" -> 12.25, "4" -> null),
"b" -> List("0" -> "R", "1" -> null, "2" -> null, "3" -> 34.75, "4" -> null),
"c" -> List("0" -> "R", "1" -> null, "2" -> null, "3" -> 56.25, "4" -> null),
"d" -> List("0" -> "R", "1" -> null, "2" -> null, "3" -> 45.65, "4" -> null))
val keys = List("a", "d")
val field = 3
m1.collect{ case (k, v) if keys.contains(k) => (k, v(field)._2) }
如果数据的顺序无关紧要,那么您可以使用Map
而不是List
,代码会稍微干净一些:
val field = "3"
m1.collect{ case (k, v) if keys.contains(k) => (k, v(field)) }
这还允许您删除
null
值,并避免使用Any
m1的类型是什么?现在它不在Scala语法中,所以代码表示有点无意义。这是元组还是JSON?这是元组。实际上,输入在csv文件中。只是从中复制了一个示例输入行。代码将从文件中读取它,并将所选的键、值作为输出。@jwvh将其转换为元组映射。val m1=Map(“…”)然后请在代码中指出。Scala元组是括号内逗号分隔的值:('c',5)
映射的键/值对通常用箭头表示,'d'->12
,但您也可以使用元组表示法,('d',12)
。